On the moon the acceleration due to gravity is $g_m = 1.62 m/s^2$. On earth, a person of mass $m = 80 kg$ manages to jump $1.4 m$. Find the height this person will reach when jumping on the moon, if the person is wearing a spacesuit with mass $m = 124 kg$.

I am a little bit confused as to whether or not the given information regarding mass is actually needed here at all. Assume that $v_0$ is equal in both jumps, and there is no rotational movement, can't we just use the formula for conservation of mechanical energy?

$$\frac{1}{2}m v_{f}^2 + mgh_f = \frac{1}{2}m v_{0}^2 + mgh_0$$

And here we can cancel the mass, $m$ since it appears in all terms. So on earth we will have:

$$gh_f = \frac{1}{2} v_{0}^2$$

$$9.8 \cdot 1.4 = \frac{1}{2} v_{0}^2$$

$$v_0 = 5.2 m/s$$

Then on the moon, since we know $v_0$, we can then find $h_f$:

$$1.62 h_f = \frac{1}{2} \cdot (5.2)^2$$

$$h_f = 8.3 m$$

Would this not be an acceptable way to solve this? If this is wrong, can anyone please explain why this is wrong conceptually?

3 Answers
3

Your solution is correct, if you assume that $v_0$ is the same on both moon and earth. The reason this seems counterintuitive is because, almost certainly, the jumper will not be able to generate the same $v_0$ on earth and moon. We can see this as followos.

The initial velocity is proportional to the integral of the force applied by the jumper's feet to the surface of the moon over time:
\begin{align}
\Delta v &= \frac{1}{m} \int_{t_i}^{t_f} dt \, F(t).
\end{align}
Assuming the force profile in time, $F(t)$ is the same on the earth and moon and is applied over the same time interval, $t_f-t_i$, we find that the initial velocity $v_0 = \Delta v$ is less on the moon owing to the larger total mass, $m_{jumper} + m_{suit}$.

Thanks a lot. This question was actually given today on a mid-term exam. However, I am taking an introductory algebraic-based physics class, so it may be quite possible that we were actually supposed to assume that $v_0$ is equal on both jumps. Is there any way to find the differences in $v_0$ without resorting to calculus?
–
user12277Oct 18 '12 at 19:35

If the statement of the problem didn't specify either the assumption that $v_0$ is the same on earth and moon or more information (about the nature of the force) then it was improperly stated. I suspect that you've interpreted it correctly. Good work and thanks for giving a complete description of your solution.
–
MarkWayneOct 18 '12 at 19:54

Thanks a lot! I am crossing my fingers I interpreted it correctly. The way I stated the problem is a direct copy of the way the problem was asked, and, as mentioned, this is a non-calculus based physics class. During the exam I thought perhaps our professor added the mass-information just to throw us a bit off track. Time will tell if I'm right :)
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user12277Oct 18 '12 at 19:59

@user12277 Often in exams, just the information you need is given. In this case you probably should've assume the amount of kinetic energy just at lift-off is the same.
–
BernhardOct 19 '12 at 5:42

I don't agree with Bernhard's suggestion that one should assume that the amount of kinetic energy is the same on earth and the moon. I'd be interested to know why he thinks this. Having equal amounts of kinetic energy at lift-off would require a lower speed, v_0 (inversely proportional to the ratio of the masses) on the moon. I don't see any physical reason or means by which one would maintain constant kinetic energy.
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MarkWayneOct 19 '12 at 16:50

Thanks a lot! As mentioned above, this is an introductory physics class with no calculus. So I am actually unsure as to whether or not we were suppose to assume $v_0$ is equal or not.
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user12277Oct 18 '12 at 19:37

Suppose we know the $g_e$ ($g$ at earth) and the man has to lift his
$c_m$ (center of mass) the same distance $s$ on the moon as on the
earth, until he loses contact with the ground. Let $h_e$ and $h_m$ be
the heigths he reaches on earth and moon respectively. Assuming the
force is the same until he loses contact to the ground, on the moon
and on earth then the work $W$ that produces during "take off" is
also the same. This work is used to bring him at maximum height.

Thank you so much for your answer. I actually got a confirmation from my professor today that this was indeed how we were supposed to interpret the problem. So it turns out I was wrong after all. Too bad, but at least now I will be more careful with problems such as these in the future when it comes to making assumptions.
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user12277Oct 19 '12 at 11:42

Yep. Work-energy will solve it. Sorry if caused any confusion. (Though the answers I gave are internally consistent.)
–
MarkWayneOct 19 '12 at 18:46